I’m working my way through College Publications’s Handbook of Deontic Logic and Normative Systems and I’ve run into a bit of trouble. I’m wondering if somebody could help explain why DD specifically, say, as opposed to KD necessitates that we regard R as a serial relation...
If a model validates the axiom DD, then for any "world" $w$ we have $w \Vdash \mathrm{OB}p \rightarrow \neg \mathrm{OB} \neg p$. We know that $w \Vdash \mathrm{OB}p \rightarrow \neg \mathrm{OB} \neg p$ holds precisely if $w \not\Vdash \mathrm{OB}p \vee w \Vdash \neg \mathrm{OB} \neg p$.
Assume that the first clause of the disjunction holds, so $\neg (w \Vdash \mathrm{OB}p)$. Expanding the definition of the satisfaction relation, we get $\neg (\forall v. Rwv \rightarrow v \Vdash p)$. By De Morgan's laws, this is equivalent to $\exists v. Rwv \wedge v \not\Vdash p$, so in this case we have $\exists v. Rwv$.
Now assume that the second clause holds, so $w \Vdash \neg \mathrm{OB} \neg p$. This means that $\neg (w \Vdash \mathrm{OB} \neg p)$ holds, and expanding the definition of Kripke satisfaction as in the previous case, we obtain $\neg (\forall v. Rwv \rightarrow v \Vdash \neg p)$. By De Morgan's laws, we have $\exists v. Rwv \wedge (v \not\Vdash \neg p)$, so again $\exists v. Rwv$.
In both cases we have at least $\exists v. Rwv$. Since this holds for any world, we must have $\forall w. \exists v. Rwv$. So to ensure the validity of the axiom DD, we must regard $R$ as a serial relation.
With respect to your second question: there are non-serial relations that validate KD (it is the usual axiom K of modal logic, which requires no frame conditions). In fact, even the empty relation validates KD.
By expanding the definition of satisfaction, we see that $w \Vdash \mathrm{OB}q$ is a sufficient condition for $w \Vdash \mathrm{OB}(p \rightarrow q) \rightarrow (\mathrm{OB}p \rightarrow \mathrm{OB}q)$.
Recall that $w \Vdash \mathrm{OB}q$ abbreviates $\forall v. Rwv \rightarrow v \Vdash q$. Notice that if $R$ is the empty relation, then $Rwv$ never holds, so the implication $Rwv \rightarrow v \Vdash q$ always holds. So the empty relation validates KD, but the empty relation is clearly not serial.